Question

Integrals, absolute value...what to do? I keep getting different answers.

Answer 1

\[\int\limits_{-2}^{0} \left| x-1 \right|\] dx=

Answer 2

the dx is supposed to be right next to the function not underneath it by the way...

Answer 3

first time i got 0, second time i got 1 and the third time i got 5/2

Answer 4

The key to understanding this problem is understnding |x-1| and its definition for x < 1. You are taking your integral where you have "redefine" or "convert" your equation first.

Answer 5

so is it redefined as -2 to -1 (x-1) plus 0 to 1 (x-1) ?

Answer 6

For x < 1, |x-1| = -(x-1) = 1-x Now you can compute your integral. You are still doing your integral at -2 to 0

Answer 7

\[\int\limits_{-2}^{0} (x-1) + \int\limits_{0}^{1} (x-1)\] so not like this?

Answer 8

the 0 supposed to be a -1

Answer 9

Now I see 2 posts with different domains. Your original looked like -2 to 0. I apologize, but I'm still unclear on what domain you are looking at.

Answer 10

sorry, so many typos... I am just unclear on how to solve using absolute values, i thought that when you have an absolute value you had to find the intercept and use that along with your domains so then you would add the integral using -2 to 0 and 0 to 1 ?? or something like that???

Answer 11

When you have an equation with absolute values and you are taking the integral, you split the integral up and have an intermediate point of where the function "changes", so you take the same lower bound for the first integral, but the upper bound becomes the "flip" x-value. Use this "flip value for the second integral lower bound and the original upper bound. Now, if I knew what your overall domain is, I could help you more.

Answer 12

I understand.
the question is number 1
http://www.math.uh.edu/~pamb/1431EMCF%2033.pdf

Answer 13

Ok, so your initial post still holds. So, just redefine your equation as y =1-x and then take your integral and you'll be fine.

Answer 14

Because in -2 to 0 for x, you will be where the definition of your function is just that, 1-x. Where x >= 1 your function is x-1, flipped around with the sign.

Answer 15

i am doing something wrong, i get 1/2 now lol. but basically im adding two "integrals" right? while using a flip x value for the second one and using the redefined equation in place of both?? sorry i am just trying to visualize this

Answer 16

I just noticed that your first post integral is different from the attachment. I'm going to work with the attachment for a few minutes and then I'll get back to you.

Answer 17

The answer to the integral is "1". I am using the attachment integral, not the one from your first post like I was using before I noticed they were different. |x+1| is -x-1 in the interval for x of (-inf, -1). |x+1| is x+1 in [-1, inf). So, you split the integral up into doing the integral on -x-1 in -2 to -1 and a second integral on x+1 in -1 to 0. The first and second integral evaluations are each 1/2. 1/2 + 1/2 = 1 for your final answer.

Answer 18

Thank you so much. I solved it again when openstudy was closed and I got 1 as well. Hooray for verification :)